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jolli1 [7]
3 years ago
12

The sum of the page numbers on the facing pages of a book is 629. what are the page numbers

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0
So basically, these numbers have to be consecutive because they're facing pages.

So you can just divide this by 2, and then add one to the second number.

629/2 = 314 R1

So the first page number is 314, and for the second one, just add one, and you get 315.
Morgarella [4.7K]3 years ago
6 0
We are looking for consecutive numbers.....x and x + 1

x + x + 1 = 629
2x + 1 = 629
2x = 629 - 1
2x = 628
x = 628/2
x = 314

x + 1 = 314 + 1 = 315

so ur 2 page numbers are 314 and 315
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Suppose X has an exponential distribution with mean equal to 23. Determine the following:
e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Mean equal to 23.

This means that m = 23, \mu = \frac{1}{23} = 0.0435

(a) P(X >10)

P(X > 10) = e^{-0.0435*10} = 0.6473

So

P(X > 10) = 0.6473

(b) P(X >20)

P(X > 20) = e^{-0.0435*20} = 0.4190

So

P(X > 20) = 0.4190

(c) P(X <30)

P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

P(X > x) = e^{-\mu x}

0.05 = e^{-0.0435x}

\ln{e^{-0.0435x}} = \ln{0.05}

-0.0435x = \ln{0.05}

x = -\frac{\ln{0.05}}{0.0435}

x = 68.87

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3 years ago
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